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36. (a) From Eq. 1328, we see that
0
/2
E
vG
M
R
=
in this problem.
Using energy
conservation, we have
1
2
mv
o
2
–
GMm/R
E
= –
GMm/r
which yields
r
= 4
R
E
/3. So the multiple of
R
E
is 4/3 or 1.33.
(b) Using the equation in the textbook immediately preceding Eq. 1328, we see that in
this problem we have
K
i
=
GMm/
2
R
E
, and the above manipulation (using energy
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Energy

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